What is the expectation of a random variable divided by an average $E\left[\frac{X_i}{\bar{X}}\right]$? Let $X_i$ be IID and $\bar{X} = \sum_{i=1}^{n} X_i$.
$$
E\left[\frac{X_i}{\bar{X}}\right] = \ ?
$$
It seems obvious, but I am having trouble formally deriving it.
 A: Let $X_1,\dots,X_n$ be independent and identically distributed random variables and define $$\bar{X}=\frac{X_1+X_2\dots+X_n}{n}.$$
Suppose that $\Pr\{\bar{X}\ne 0\}=1$. Since the $X_i$'s are identically distributed, symmetry tells us that, for $i=1,\dots n$, the (dependent) random variables $X_i/\bar{X}$ have the same distribution:
$$
  \frac{X_1}{\bar{X}} \sim \frac{X_2}{\bar{X}} \sim \dots \sim \frac{X_n}{\bar{X}}.
$$
If the expectations $\mathrm{E}[X_i/\bar{X}]$ exist (this is a crucial point), then
$$
\mathrm{E}\left[ \frac{X_1}{\bar{X}} \right] = \mathrm{E}\left[ \frac{X_2}{\bar{X}} \right] = \dots = \mathrm{E}\left[ \frac{X_n}{\bar{X}} \right],
$$
and, for $i=1,\dots,n$, we have
$$
\begin{align}
\mathrm{E}\left[ \frac{X_i}{\bar{X}} \right] &= \frac{1}{n} \left( \mathrm{E}\left[ \frac{X_1}{\bar{X}} \right] + \mathrm{E}\left[ \frac{X_2}{\bar{X}} \right] + \dots + \mathrm{E}\left[ \frac{X_n}{\bar{X}} \right] \right) \\
&= \frac{1}{n}\,\mathrm{E}\left[ \frac{X_1}{\bar{X}} + \frac{X_2}{\bar{X}} + \dots + \frac{X_n}{\bar{X}} \right] \\
&= \frac{1}{n}\,\mathrm{E}\left[ \frac{X_1+X_2+\dots+X_n}{\bar{X}} \right] \\ 
&= \frac{1}{n}\,\mathrm{E}\left[ \frac{n\bar{X}}{\bar{X}} \right] \\
&= \frac{n}{n}\,\mathrm{E}\left[ \frac{\bar{X}}{\bar{X}} \right] = 1.
\end{align}
$$
Let's see if we can check this by simple Monte Carlo.
x <- matrix(rgamma(10^6, 1, 1), nrow = 10^5)
mean(x[, 3] / rowMeans(x))

[1] 1.00511

Fine, and the results don't change much under repetition.
